Firstly, label each coefficient: coefficient in front of x2 “a”, is equal to 4. “b” is the coefficient in front of x is equal to +4 “c” is equal to -15

The procedure for factorising starts with working out the grouping values: We want two numbers that multiply to get, “a” * ”c”= 4*-15 = -60 And adds to equal “b” = 4

Try combinations of numbers that multiply to get -60 and add to get 4. Try -15 and 4: -154 equals -60, however adds to get -11, so not the combination Try 10 and -6: 10-6 equals -60 and adds to get a difference of 4, so is the combination.

Now we split the middle of term of the quadratic, using the combination: 4x can be rewritten as 10x-6x, therefore 4x2+10x -6x-15

Now group the terms and factorise The first group is the first part of the quadratic: 4x2+10x, this can be factorised into 2x(2x+5) The second group is -6x-15, this can be factorised into -3(2x+5) So 4x2+10x -6x-15 can written as: 2x(2x+5)-3(2x+5). You can see 2x+5 is common, so you can factorise that out, hence you are left with the answer: (2x+5)(2x-3)

(3x+9) (7x+5)

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